Problem: In the diagram shown, $\overrightarrow{OA}\perp\overrightarrow{OC}$ and $\overrightarrow{OB}\perp\overrightarrow{OD}$. If $\angle{AOD}$ is 3.5 times $\angle{BOC}$, what is $\angle{AOD}$? [asy]
unitsize(1.5cm);
defaultpen(linewidth(.7pt)+fontsize(10pt));
dotfactor=4;

pair O=(0,0), A=dir(0), B=dir(50), C=dir(90), D=dir(140);
pair[] dots={O,A,B,C,D};

dot(dots);
draw(O--1.2*D,EndArrow(4));
draw(O--1.2*B,EndArrow(4));
draw(O--1.2*C,EndArrow(4));
draw(O--1.2*A,EndArrow(4));

label("$D$",D,SW);
label("$C$",C,W);
label("$B$",B,E);
label("$A$",A,N);
label("$O$",O,S);
[/asy]
Let $x$ denote the measure in degrees of $\angle BOC$.  Because $\angle BOD$ and $\angle COA$ are right angles, $\angle COD$ and $\angle BOA$ each measure $90-x$ degrees.  Therefore $\angle AOD=x+(90-x)+(90-x)$ degrees.  Solving \[
3.5x=x+90-x+90-x
\]we find $x=180/4.5=40$.  Therefore $\angle AOD=180^\circ-40^\circ=\boxed{140\text{ degrees}}$.